1
00:00:02,010 --> 00:00:06,696
By now you have had a lot
of opportunity to practice

2
00:00:06,696 --> 00:00:07,974
differentiating common
functions.

3
00:00:09,060 --> 00:00:12,124
And you have learned many
techniques of Differentiation.

4
00:00:12,980 --> 00:00:16,685
So for example, if I were to
give you a function, let's call

5
00:00:16,685 --> 00:00:18,395
this function capital F of X.

6
00:00:21,150 --> 00:00:23,510
If it's a fairly simple
function, you'll know

7
00:00:23,510 --> 00:00:24,690
how to differentiate it.

8
00:00:34,760 --> 00:00:39,556
And so by differentiating it,
you'll be able to calculate its

9
00:00:39,556 --> 00:00:43,916
derivative, which you'll
remember we denote as DF DX. So

10
00:00:43,916 --> 00:00:47,840
that's a process that you're
very familiar with already.

11
00:00:48,860 --> 00:00:54,908
Now in this unit we're going to
refer to DF DX as little F of X.

12
00:00:57,620 --> 00:00:59,490
So little F of X.

13
00:01:00,210 --> 00:01:04,148
Is the derivative of big
F of X. Let me write

14
00:01:04,148 --> 00:01:06,296
that down little F of X.

15
00:01:07,830 --> 00:01:09,618
Is the derivative.

16
00:01:18,240 --> 00:01:19,488
As big banks.

17
00:01:20,800 --> 00:01:24,652
And as I say, that's a process
that you're very familiar with.

18
00:01:24,652 --> 00:01:29,146
What we want to do now is try to
work this process in reverse.

19
00:01:29,146 --> 00:01:32,677
Work it backwards. In other
words, we want to start with

20
00:01:32,677 --> 00:01:36,499
little F. And try and
come back this route.

21
00:01:38,810 --> 00:01:42,890
And try and find the function or
functions capital F which when

22
00:01:42,890 --> 00:01:46,290
they are differentiated will
give you little Earth. So we

23
00:01:46,290 --> 00:01:50,710
want to carry out this process
in reverse. We can think of this

24
00:01:50,710 --> 00:01:51,730
as anti differentiation.

25
00:01:59,050 --> 00:02:03,091
So we can think of this anti
differentiation as

26
00:02:03,091 --> 00:02:04,438
differentiation in reverse.

27
00:02:05,770 --> 00:02:10,580
So big F of X we're going to
call the anti derivative of

28
00:02:10,580 --> 00:02:12,060
little F of X.

29
00:02:26,860 --> 00:02:30,584
So by these two results in mind
and try not to get confused with

30
00:02:30,584 --> 00:02:34,308
a capital F in the lower case F
little F is the derivative of

31
00:02:34,308 --> 00:02:40,010
Big F. Because little F
is DF DX.

32
00:02:41,650 --> 00:02:45,420
And big F is the anti derivative
of little laugh.

33
00:02:47,200 --> 00:02:50,984
Now you may have already looked
at a previous video called

34
00:02:50,984 --> 00:02:52,016
integration as summation.

35
00:02:53,120 --> 00:02:57,560
And in that video, the concept
of an integral is defined in

36
00:02:57,560 --> 00:03:02,000
terms of the sum of lots of
rectangular areas under a curve.

37
00:03:02,940 --> 00:03:07,113
To calculate an integral, you
need to find the limit of a son.

38
00:03:07,700 --> 00:03:10,724
And that's a very cumbersome
and impractical process. What

39
00:03:10,724 --> 00:03:14,756
we're going to learn about in
this video is how to find

40
00:03:14,756 --> 00:03:18,452
integrals, not by finding the
limit of a sum, but instead

41
00:03:18,452 --> 00:03:19,460
by using antiderivatives.

42
00:03:20,520 --> 00:03:22,668
Let's start off
with the example.

43
00:03:24,110 --> 00:03:27,971
Suppose we start off with the
function capital F of X.

44
00:03:29,660 --> 00:03:34,084
Is equal to three
X squared plus 7X.

45
00:03:36,310 --> 00:03:37,190
Minus 2.

46
00:03:38,680 --> 00:03:41,364
And what I'm going to do is
something that you're already

47
00:03:41,364 --> 00:03:43,316
very familiar with. We're
going to differentiate it.

48
00:03:44,720 --> 00:03:46,350
And if we differentiate it.

49
00:03:47,290 --> 00:03:50,800
We get TF DX equals.

50
00:03:51,650 --> 00:03:55,355
Turn by turn the derivative of
three X squared is going to be.

51
00:03:55,940 --> 00:03:57,470
2306 X.

52
00:04:00,040 --> 00:04:03,232
And the derivative of Seven X is
just going to be 7.

53
00:04:05,230 --> 00:04:09,179
What about the minus two? Well,
you remember that the derivative

54
00:04:09,179 --> 00:04:13,487
of a constant is 0, so when we
do this differentiation process,

55
00:04:13,487 --> 00:04:14,923
the minus two disappears.

56
00:04:15,490 --> 00:04:19,310
So our derivative DF DX is just
six X +7.

57
00:04:20,770 --> 00:04:25,450
So I'm going to write little F
of X is 6 X +7.

58
00:04:26,980 --> 00:04:31,336
And think about what will happen
when we reverse the process when

59
00:04:31,336 --> 00:04:32,788
we do anti differentiation.

60
00:04:33,920 --> 00:04:38,419
Ask yourself what is an anti
derivative of 6X plus 7?

61
00:04:56,100 --> 00:04:59,565
Now we already have an answer to
this question and anti

62
00:04:59,565 --> 00:05:01,140
derivative of 6X plus 7.

63
00:05:02,150 --> 00:05:05,879
Is 3 X squared plus 7X minus
two, so the answer.

64
00:05:07,530 --> 00:05:13,755
Is F of X is 3 X squared plus
7X minus two, but I'm afraid

65
00:05:13,755 --> 00:05:18,320
that's not the whole story.
We've already seen that when we

66
00:05:18,320 --> 00:05:22,470
differentiate it, three X
squared plus 7X minus two, the

67
00:05:22,470 --> 00:05:23,715
minus two disappeared.

68
00:05:24,790 --> 00:05:28,409
In other words, whatever number
had been in here, whether that

69
00:05:28,409 --> 00:05:33,015
minus two had been minus 8 or
plus 10 or 0, whatever it would

70
00:05:33,015 --> 00:05:35,647
still have disappeared. We've
lost some information during

71
00:05:35,647 --> 00:05:39,266
this process of Differentiation,
and when we want to reverse it

72
00:05:39,266 --> 00:05:43,543
when we want to start with the
six, XX have Seven, and working

73
00:05:43,543 --> 00:05:47,162
backwards, we've really no idea
what that minus two might have

74
00:05:47,162 --> 00:05:49,136
been. It could have been another

75
00:05:49,136 --> 00:05:54,382
number. So this leads us to the
conclusion that once we found an

76
00:05:54,382 --> 00:05:57,892
antiderivative like this one,
three X squared plus, 7X minus

77
00:05:57,892 --> 00:06:02,530
2. Than any constant added
onto this will still be an

78
00:06:02,530 --> 00:06:03,270
anti derivative.

79
00:06:05,790 --> 00:06:07,110
If F of X.

80
00:06:08,460 --> 00:06:10,768
Is an anti derivative.

81
00:06:16,640 --> 00:06:18,180
A little F of X.

82
00:06:19,260 --> 00:06:20,559
Then so too.

83
00:06:24,070 --> 00:06:28,737
Is F of X the anti derivative
we've found plus any constant at

84
00:06:28,737 --> 00:06:33,045
all we choose. See is what we
call an arbitrary constant. Any

85
00:06:33,045 --> 00:06:37,712
value of see any constant we can
add on to the anti derivative

86
00:06:37,712 --> 00:06:41,302
we've found and then we've got
another antiderivative. So in

87
00:06:41,302 --> 00:06:45,251
fact there are lots and lots of
antiderivatives of 6X plus

88
00:06:45,251 --> 00:06:49,200
Seven, just as another example,
we could have had over here.

89
00:06:50,860 --> 00:06:53,940
Three X squared plus 7X minus 8.

90
00:06:55,590 --> 00:07:00,510
Or even just three X squared
plus 7X where the constant term

91
00:07:00,510 --> 00:07:05,020
was zero. So lots and lots of
different antiderivatives for a

92
00:07:05,020 --> 00:07:06,660
single term over here.

93
00:07:09,120 --> 00:07:10,568
Let's do another example.

94
00:07:13,450 --> 00:07:14,950
Suppose this time we look at.

95
00:07:15,940 --> 00:07:21,220
A cubic function that suppose F
of X is 4X cubed.

96
00:07:23,230 --> 00:07:24,998
Minus Seven X squared.

97
00:07:26,440 --> 00:07:27,640
Plus 12X

98
00:07:29,180 --> 00:07:29,930
minus 4.

99
00:07:32,180 --> 00:07:34,772
Again, you know how to
differentiate this. You've

100
00:07:34,772 --> 00:07:37,040
had a lot of practice
differentiating functions

101
00:07:37,040 --> 00:07:37,688
like this.

102
00:07:39,220 --> 00:07:42,748
So the derivative DF
DX is going to be.

103
00:07:43,770 --> 00:07:45,510
Three 412 X squared.

104
00:07:48,710 --> 00:07:52,730
The derivative of minus
Seven X squared is going to

105
00:07:52,730 --> 00:07:53,936
be minus 14X.

106
00:07:56,540 --> 00:07:58,970
And the derivative
of 12 X is just 12.

107
00:08:01,900 --> 00:08:08,380
So think of our little F of X as
being the function 12 X squared

108
00:08:08,380 --> 00:08:10,108
minus 14X at 12.

109
00:08:12,470 --> 00:08:16,930
Notice again the point that
the minus four in the

110
00:08:16,930 --> 00:08:18,268
differentiation process
disappears.

111
00:08:19,840 --> 00:08:23,470
So now ask the question, what's
an anti derivative of this

112
00:08:23,470 --> 00:08:24,790
function here little F.

113
00:08:25,630 --> 00:08:29,098
Well, we've got one answer. It's
on the page already. It's this

114
00:08:29,098 --> 00:08:31,121
big F is an anti derivative of

115
00:08:31,121 --> 00:08:35,690
Little F. But we've seen
that any constant added on

116
00:08:35,690 --> 00:08:38,385
here will still yield
another antiderivative, so

117
00:08:38,385 --> 00:08:42,235
we can write down lots of
other ones, for example.

118
00:08:44,300 --> 00:08:46,256
If we add on a constant.

119
00:08:48,460 --> 00:08:51,448
And let's suppose we add on the
constant 10 onto this one.

120
00:08:52,120 --> 00:08:57,862
Then we'll get the function 4X
cubed, minus Seven X squared.

121
00:08:58,800 --> 00:09:03,720
Plus 12X plus six is also an
anti derivative of Little F.

122
00:09:04,890 --> 00:09:08,166
And the same goes by adding
adding on any other constant at

123
00:09:08,166 --> 00:09:11,169
all we choose. If we add on
minus six to this.

124
00:09:14,910 --> 00:09:19,564
Will get 4X cubed, minus Seven X
squared plus 12X and adding on

125
00:09:19,564 --> 00:09:23,860
minus six or just leave us this
so that that is another

126
00:09:23,860 --> 00:09:26,008
antiderivative. There's an
infinite number of

127
00:09:26,008 --> 00:09:27,440
antiderivatives of Little F.

128
00:09:28,810 --> 00:09:30,262
Now in all the examples we've

129
00:09:30,262 --> 00:09:34,250
looked at. I've in this in a
sense, giving you the answer

130
00:09:34,250 --> 00:09:37,451
right at the beginning because
we started with big F, we

131
00:09:37,451 --> 00:09:41,234
differentiated it to find little
F, so we knew the answer all the

132
00:09:41,234 --> 00:09:44,726
time. In practice, you won't
know the answer all the time, so

133
00:09:44,726 --> 00:09:47,927
one way that we can help
ourselves by referring to a

134
00:09:47,927 --> 00:09:50,546
table of antiderivatives. Now a
table of antiderivatives will

135
00:09:50,546 --> 00:09:52,001
look something like this one.

136
00:09:53,000 --> 00:09:56,888
There's a copy of this in the
notes accompanying the video and

137
00:09:56,888 --> 00:09:59,804
a table of antiderivatives were
list lots of functions.

138
00:10:00,160 --> 00:10:00,570
F.

139
00:10:01,720 --> 00:10:03,740
And then the anti derivative.

140
00:10:04,390 --> 00:10:08,251
Big F in the next column and
you'll see every anti

141
00:10:08,251 --> 00:10:12,463
derivative will have a plus.
See attached to it where C is

142
00:10:12,463 --> 00:10:14,920
an arbitrary constant. Any
constant you choose.

143
00:10:23,820 --> 00:10:28,708
What I want to do now is I
want to link the concept

144
00:10:28,708 --> 00:10:31,340
of an antiderivative with
integration as summation.

145
00:10:33,210 --> 00:10:36,060
Let's think of a function.

146
00:10:38,960 --> 00:10:40,840
Y equals F of X.

147
00:10:46,040 --> 00:10:48,790
Let's suppose the graph of the
function looks like this.

148
00:10:51,000 --> 00:10:53,664
I'm going to consider
functions which lie entirely

149
00:10:53,664 --> 00:10:56,994
above the X axis, so I'm
going to restrict our

150
00:10:56,994 --> 00:11:00,657
attention to the region above
the X axis were looking up

151
00:11:00,657 --> 00:11:00,990
here.

152
00:11:02,240 --> 00:11:05,906
And also I want to restrict
attention to the right of the Y

153
00:11:05,906 --> 00:11:09,008
axis, so I'm looking to the
right of this line here.

154
00:11:11,550 --> 00:11:16,750
Let's ask ourselves what is the
area under the graph of Y equals

155
00:11:16,750 --> 00:11:17,950
F of X?

156
00:11:19,650 --> 00:11:23,563
Well, surely that depends on
how far I want to move to the

157
00:11:23,563 --> 00:11:26,272
right hand side. Let's
suppose I want to move.

158
00:11:28,060 --> 00:11:31,030
To the place where X
has an X Coordinate X.

159
00:11:33,590 --> 00:11:35,557
Then clearly, if X is a large

160
00:11:35,557 --> 00:11:39,118
number. The area under this
graph here will be large,

161
00:11:39,118 --> 00:11:42,440
whereas if X is a small
number, the area under the

162
00:11:42,440 --> 00:11:45,460
graph will be quite small
indeed. Effects is actually 0,

163
00:11:45,460 --> 00:11:49,386
so this line is actually lying
on the Y axis than the area

164
00:11:49,386 --> 00:11:51,198
under the graph will be 0.

165
00:11:52,830 --> 00:11:55,974
I'm going to do note the
area under the graph by a,

166
00:11:55,974 --> 00:11:57,284
so A is the area.

167
00:12:01,680 --> 00:12:02,270
Under

168
00:12:04,050 --> 00:12:05,240
why is F of X?

169
00:12:07,490 --> 00:12:11,715
And as I've just said, the area
will depend upon the value that

170
00:12:11,715 --> 00:12:15,940
we choose for X. In other
words, A is a function of XA

171
00:12:15,940 --> 00:12:20,815
depends upon X, so we write
that like this, a is a of X and

172
00:12:20,815 --> 00:12:24,715
that shows the dependence of
the area on the value we choose

173
00:12:24,715 --> 00:12:28,940
for X. As I said, LG X larger
area small acts smaller area.

174
00:12:30,600 --> 00:12:34,718
Ask yourself. What is the height
of this line here?

175
00:12:37,320 --> 00:12:38,928
Well, this point here.

176
00:12:39,590 --> 00:12:42,995
Lies. On the curve Y equals F of

177
00:12:42,995 --> 00:12:48,426
X. So the Y value at that point
is simply the function evaluated

178
00:12:48,426 --> 00:12:49,858
at this X value.

179
00:12:50,580 --> 00:12:53,127
So the Y value here is just F of

180
00:12:53,127 --> 00:12:57,440
X. In other words, the
height of this line or the

181
00:12:57,440 --> 00:13:01,136
length of this line is just
F of X. The function

182
00:13:01,136 --> 00:13:02,480
evaluated at this point.

183
00:13:04,940 --> 00:13:08,678
Now I'd like you to think what
happens if we just increase X by

184
00:13:08,678 --> 00:13:09,746
a very small amount.

185
00:13:10,340 --> 00:13:13,028
So I want to just move this
point a little bit further

186
00:13:13,028 --> 00:13:13,700
to the right.

187
00:13:14,960 --> 00:13:16,120
And see what happens.

188
00:13:17,110 --> 00:13:21,127
I'm going to increase X by a
little bit and that little bit

189
00:13:21,127 --> 00:13:25,144
that distance in here. I'm going
to call Delta X. Delta X stands

190
00:13:25,144 --> 00:13:29,779
for a small change in X or we
call it a small increment in X.

191
00:13:31,930 --> 00:13:34,100
By increasing axle little bit.

192
00:13:34,890 --> 00:13:37,919
What we're doing is we're adding
a little bit more to this area.

193
00:13:39,720 --> 00:13:42,447
This is the additional
contribution to the area. This

194
00:13:42,447 --> 00:13:43,659
shaded region in here.

195
00:13:44,800 --> 00:13:47,738
And that's a little bit extra
area, so I'm going to call that.

196
00:13:49,830 --> 00:13:53,726
Delta A, That's an incrementing
area changing area.

197
00:13:55,630 --> 00:13:57,772
I'm going to write down
an expression for Delta

198
00:13:57,772 --> 00:13:59,200
a try and work it out.

199
00:14:01,940 --> 00:14:06,000
Let's try and see what it is,
but I can't get it exactly. But

200
00:14:06,000 --> 00:14:09,770
if I note that the height of
this line is F of X.

201
00:14:10,500 --> 00:14:14,655
And the width of this column in
here is Delta X. Then I can get

202
00:14:14,655 --> 00:14:17,702
an approximate value for this
area by assuming that it's a

203
00:14:17,702 --> 00:14:20,472
rectangular section. In other
words, I'm going to ignore this

204
00:14:20,472 --> 00:14:22,411
little bit at the top in there.

205
00:14:23,340 --> 00:14:27,144
If I assume it's a rectangular
section, then the area here this

206
00:14:27,144 --> 00:14:28,412
additional area Delta A.

207
00:14:29,320 --> 00:14:32,648
Is F of X multiplied by Delta X.

208
00:14:35,980 --> 00:14:39,030
Now, that's only approximately
true, so this is really an

209
00:14:39,030 --> 00:14:40,250
approximately equal to symbol.

210
00:14:41,170 --> 00:14:43,487
This additional area is
approximately equal to

211
00:14:43,487 --> 00:14:45,473
F of X Times Delta X.

212
00:14:48,210 --> 00:14:53,922
Let me now divide both sides by
Delta X. That's going to give me

213
00:14:53,922 --> 00:14:58,818
Delta a over Delta. X is
approximately equal to F of X.

214
00:15:01,060 --> 00:15:04,240
How can we make it more
accurate? Well, one way we can

215
00:15:04,240 --> 00:15:07,420
make this more accurate is by
choosing this column to be even

216
00:15:07,420 --> 00:15:10,070
thinner by letting Delta X be
smaller, because then this

217
00:15:10,070 --> 00:15:12,720
additional contribution that
I've got in here that I didn't

218
00:15:12,720 --> 00:15:16,430
count is reducing its reducing
in size. So what I want to do is

219
00:15:16,430 --> 00:15:20,140
I want to let Delta X get even
smaller and smaller and in the

220
00:15:20,140 --> 00:15:21,995
end I want to take the limit.

221
00:15:22,990 --> 00:15:27,863
As Delta X tends to zero of
Delta over Delta X.

222
00:15:28,420 --> 00:15:31,440
And when I do that, this
approximation in here will

223
00:15:31,440 --> 00:15:34,460
become exact and that will give
us F of X.

224
00:15:36,620 --> 00:15:40,670
Now, if you've studied the unit
on differentiation from first

225
00:15:40,670 --> 00:15:44,720
principles, you realize that
this in here is the definition

226
00:15:44,720 --> 00:15:49,985
of the derivative of A with
respect to X, which we write as

227
00:15:49,985 --> 00:15:55,655
DADX. So we have the result that
DADX is a little F of X.

228
00:15:58,540 --> 00:16:00,108
Let's explore this a
little bit further.

229
00:16:03,830 --> 00:16:09,020
We have the ADX is a little
F of X.

230
00:16:10,480 --> 00:16:13,984
What does this mean? Well,
it means that little laugh

231
00:16:13,984 --> 00:16:15,604
is the derivative of A.

232
00:16:26,290 --> 00:16:29,995
But it also means if we think
back to the discussion that we

233
00:16:29,995 --> 00:16:33,700
had at the beginning of this
video, that a must be the anti

234
00:16:33,700 --> 00:16:38,440
derivative of F. A is
an anti derivative.

235
00:16:45,740 --> 00:16:53,328
F. In other words, the area is
F of X. The anti derivative of

236
00:16:53,328 --> 00:16:56,038
little F was any constant.

237
00:16:58,970 --> 00:17:02,642
That's going to be an important
result. What it's saying is that

238
00:17:02,642 --> 00:17:07,538
if we have a function, why is F
of X and we want to find the

239
00:17:07,538 --> 00:17:11,210
area under the graph? What we do
is we calculate an anti

240
00:17:11,210 --> 00:17:13,046
derivative of Little F which is

241
00:17:13,046 --> 00:17:15,578
big F. And use this expression.

242
00:17:16,200 --> 00:17:17,480
To find the area.

243
00:17:18,520 --> 00:17:21,303
There's one thing we don't know
in this expression at the

244
00:17:21,303 --> 00:17:23,580
moment, and it's this CC.
Remember, is an arbitrary

245
00:17:23,580 --> 00:17:27,122
constant, but we can get a value
for. See if we just look back

246
00:17:27,122 --> 00:17:28,640
again. So the graph I drew.

247
00:17:30,340 --> 00:17:34,370
And ask yourself what will be
the area under this curve when X

248
00:17:34,370 --> 00:17:38,400
is chosen to be zero? Well, if
you remember, we said that if

249
00:17:38,400 --> 00:17:41,500
this vertical line here had been
on the Y axis.

250
00:17:42,340 --> 00:17:44,490
Then the area under the
curve would have been 0.

251
00:17:45,580 --> 00:17:48,570
So this gives us a
condition. It tells us that

252
00:17:48,570 --> 00:17:50,962
when X is 0, the area is 0.

253
00:17:56,890 --> 00:17:58,758
When X is 0.

254
00:18:00,240 --> 00:18:04,848
Area is 0. What does this mean?
While the area being 0?

255
00:18:06,550 --> 00:18:07,858
X being 0.

256
00:18:09,720 --> 00:18:13,680
Plus a constant and this
condition then gives us a value

257
00:18:13,680 --> 00:18:18,000
for C, so C must be equal to
minus F of North.

258
00:18:20,690 --> 00:18:26,015
And that value for C and go back
in this result here. So we have

259
00:18:26,015 --> 00:18:30,630
the final result that the area
under the graph is given by big

260
00:18:30,630 --> 00:18:33,470
F of X minus big F of note.

261
00:18:40,810 --> 00:18:42,364
Let me just write that
down again.

262
00:18:51,140 --> 00:18:51,560
OK.

263
00:18:55,670 --> 00:18:59,861
Now let's look at this problem.
Supposing that I'm interested in

264
00:18:59,861 --> 00:19:04,814
finding the area under the graph
of Y is little F of X.

265
00:19:08,460 --> 00:19:09,628
Up to the point.

266
00:19:11,530 --> 00:19:13,696
Where X has the value be.

267
00:19:14,640 --> 00:19:19,736
So I want the area from the Y
axis, which is where we're

268
00:19:19,736 --> 00:19:23,656
working from above the X axis up
to the point.

269
00:19:24,280 --> 00:19:25,480
Where X equals B.

270
00:19:26,480 --> 00:19:28,650
We can use this boxed
formula here.

271
00:19:30,070 --> 00:19:34,450
When X is be will get a of B.
That's the area up to be.

272
00:19:36,500 --> 00:19:38,630
Will be F of B.

273
00:19:41,320 --> 00:19:43,940
Minus F of note.

274
00:19:45,420 --> 00:19:48,907
So this expression will give you
the area up to be.

275
00:19:58,850 --> 00:20:00,110
Suppose now I want.

276
00:20:00,180 --> 00:20:03,744
Area up to A and let's
Suppose A is about here.

277
00:20:04,840 --> 00:20:06,008
So now I'm interested.

278
00:20:08,080 --> 00:20:13,360
In this area in here, which is
the area from the Y axis up to a

279
00:20:13,360 --> 00:20:17,320
well, again using the same
formula, the area up to a which

280
00:20:17,320 --> 00:20:18,640
is a of A.

281
00:20:20,150 --> 00:20:23,770
Is F evaluated at the X value
which is a?

282
00:20:26,470 --> 00:20:29,328
Subtract. Big F of note.

283
00:20:31,770 --> 00:20:34,626
So I have two expressions on
the page here, one for the

284
00:20:34,626 --> 00:20:35,578
area up to be.

285
00:20:37,360 --> 00:20:38,936
And one for the area up to A.

286
00:20:40,660 --> 00:20:45,764
Ask yourself How do I find
this area in here? That's

287
00:20:45,764 --> 00:20:47,620
the area between A&B.

288
00:20:49,160 --> 00:20:54,284
Well, the area between A&B we
can think of as the area up to

289
00:20:54,284 --> 00:20:57,576
be. Subtract the area up to A.

290
00:20:58,820 --> 00:21:02,576
So if we find the difference of
these two quantities, that will

291
00:21:02,576 --> 00:21:04,454
give us the area between A&B.

292
00:21:05,330 --> 00:21:06,790
So we get the area.

293
00:21:09,730 --> 00:21:11,910
Under Wise F of X.

294
00:21:14,960 --> 00:21:18,782
From X equals A to X equals

295
00:21:18,782 --> 00:21:23,460
B. Is the area under
this graph from X is A

296
00:21:23,460 --> 00:21:25,847
to X is B is given by?

297
00:21:26,890 --> 00:21:29,902
Once the area up to be, subtract
the area up to A.

298
00:21:30,520 --> 00:21:33,190
So if we just find the
difference of these two

299
00:21:33,190 --> 00:21:35,860
quantities will have F of B
minus F of A.

300
00:21:40,810 --> 00:21:45,217
Add minus F of not minus minus F
of notes. These terms will

301
00:21:45,217 --> 00:21:49,285
cancel out. So In other words
that's the result we need. The

302
00:21:49,285 --> 00:21:51,997
area under this graph between
A&B is just.

303
00:21:52,610 --> 00:21:54,808
Big F of B minus big FA.

304
00:21:59,570 --> 00:22:01,516
Let me just write that
down again.

305
00:22:26,380 --> 00:22:30,892
The area and Y equals F of X
between ex is an ex is B is

306
00:22:30,892 --> 00:22:35,686
given by big F of B minus big F
of a where big F remember is an

307
00:22:35,686 --> 00:22:37,096
anti derivative of Little F.

308
00:22:42,050 --> 00:22:43,298
Any answer derivative?

309
00:22:54,450 --> 00:22:57,570
So this is a very important
result because it means that if

310
00:22:57,570 --> 00:22:59,390
I give you a function little F

311
00:22:59,390 --> 00:23:04,117
of X. And you can calculate an
anti derivative of its big F.

312
00:23:04,820 --> 00:23:08,472
And you can find the area
under the graph merely by

313
00:23:08,472 --> 00:23:10,796
evaluating that anti
derivative at B, evaluating

314
00:23:10,796 --> 00:23:13,784
it A and finding the
difference of the two

315
00:23:13,784 --> 00:23:14,116
quantities.

316
00:23:16,470 --> 00:23:19,590
Now in the previous video on
integration by summation, the

317
00:23:19,590 --> 00:23:23,022
area under under a graph was
found in a slightly different

318
00:23:23,022 --> 00:23:24,582
way. What was done there?

319
00:23:25,200 --> 00:23:30,876
Was that the area under the
graph between A&B was found by

320
00:23:30,876 --> 00:23:35,133
dividing the area into lots of
thin rectangular strips?

321
00:23:36,330 --> 00:23:39,582
Finding the area of each of the
rectangles and adding them all

322
00:23:39,582 --> 00:23:45,312
up. And doing that we had this
result that the area under the

323
00:23:45,312 --> 00:23:50,200
graph between A&B was given by
the limit as Delta X tends to

324
00:23:50,200 --> 00:23:54,712
zero of the sum of all these
rectangular areas, which was F

325
00:23:54,712 --> 00:24:00,728
of X Delta X between X is A and
X is be. If you're not familiar

326
00:24:00,728 --> 00:24:05,992
with that, I would advise you go
back and have a look at that

327
00:24:05,992 --> 00:24:09,376
other video on integration by
summation, but that formula.

328
00:24:10,200 --> 00:24:13,144
For this area was derived
in that video.

329
00:24:14,190 --> 00:24:17,536
And this formula
defines what we mean

330
00:24:17,536 --> 00:24:21,360
by the definite
integral from A to B

331
00:24:21,360 --> 00:24:23,750
of F of X DX.

332
00:24:25,400 --> 00:24:28,130
That's how we defined
a definite integral.

333
00:24:29,400 --> 00:24:32,436
And if we wanted to define a
definite integral, if we wanted

334
00:24:32,436 --> 00:24:35,725
to calculate a definite integral
in that video, we had to do it

335
00:24:35,725 --> 00:24:38,255
through the process of finding
the limit of a sum.

336
00:24:39,780 --> 00:24:43,784
Now the the process of finding
the limit of a sum is quite

337
00:24:43,784 --> 00:24:47,256
cumbersome and impractical. But
what we've learned now is that

338
00:24:47,256 --> 00:24:51,426
we don't have to use the limit
of a sum to find the area. We

339
00:24:51,426 --> 00:24:52,816
can use these anti derivatives.

340
00:24:54,120 --> 00:24:56,910
Because we've got two
expressions for the area, we've

341
00:24:56,910 --> 00:25:00,567
got this expression. As a
definite integral, and we've got

342
00:25:00,567 --> 00:25:03,377
this expression in terms of
antiderivatives, and if we put

343
00:25:03,377 --> 00:25:05,063
all that together will end up

344
00:25:05,063 --> 00:25:09,056
with this result. The definite
integral from A to B.

345
00:25:10,430 --> 00:25:11,810
F of X DX.

346
00:25:13,180 --> 00:25:14,470
Is FB.

347
00:25:16,010 --> 00:25:17,898
Minus F of A.

348
00:25:21,860 --> 00:25:26,337
In other words, the definite
integral of little F between the

349
00:25:26,337 --> 00:25:30,000
limits of A&B is found by
evaluating this expression,

350
00:25:30,000 --> 00:25:34,070
where big F is an anti
derivative of Little F.

351
00:25:39,800 --> 00:25:41,606
Let me write that formula
down again.

352
00:25:43,270 --> 00:25:45,208
The integral from A to B.

353
00:25:46,710 --> 00:25:54,088
F of X DX is given by
big FB minus big F of A.

354
00:25:55,860 --> 00:25:56,946
Let me give you an example.

355
00:26:02,130 --> 00:26:06,010
Suppose we're interested
in the problem of finding

356
00:26:06,010 --> 00:26:10,860
the area under the graph
of Y equals X squared.

357
00:26:14,520 --> 00:26:17,352
And let's suppose for the sake
of argument we want the area

358
00:26:17,352 --> 00:26:21,820
under the graph. Between X is
not an ex is one, so we're

359
00:26:21,820 --> 00:26:23,024
interested in this area.

360
00:26:26,030 --> 00:26:29,594
In the previous unit on
integration as a summation,

361
00:26:29,594 --> 00:26:33,950
this was done by dividing
this area into lots of thin

362
00:26:33,950 --> 00:26:34,742
rectangular strips.

363
00:26:36,040 --> 00:26:40,360
And finding the area of each of
those rectangles separately.

364
00:26:40,980 --> 00:26:42,186
And then adding them all up.

365
00:26:43,080 --> 00:26:47,580
That gave rise to this formula
that the area is the limit.

366
00:26:48,660 --> 00:26:50,748
As Delta X tends to 0.

367
00:26:51,900 --> 00:26:57,257
Of the sum from X equals not to
X equals 1.

368
00:26:57,940 --> 00:27:00,270
Of X squared Delta X.

369
00:27:01,460 --> 00:27:04,771
So the area was expressed as the
limit of a sum.

370
00:27:05,830 --> 00:27:11,794
And in turn, that defines the
definite integral. X is not to

371
00:27:11,794 --> 00:27:14,279
one of X squared DX.

372
00:27:16,360 --> 00:27:20,380
Now open till now if you
wanted to work this area out

373
00:27:20,380 --> 00:27:24,735
the way you would do it would
be by finding the limit of

374
00:27:24,735 --> 00:27:27,080
this some, but that's
impractical and cumbersome.

375
00:27:27,080 --> 00:27:29,760
Instead, we're going to use
this result using

376
00:27:29,760 --> 00:27:30,095
antiderivatives.

377
00:27:31,780 --> 00:27:35,344
Our little F of X in this
case is X squared.

378
00:27:40,350 --> 00:27:44,508
And this formula at the top of
the page tells us that we can

379
00:27:44,508 --> 00:27:46,587
evaluate this definite
integral by finding an

380
00:27:46,587 --> 00:27:49,557
antiderivative capital F. So
we want to do that first.

381
00:27:49,557 --> 00:27:53,418
Well, if little F is X
squared, big F of X, well we

382
00:27:53,418 --> 00:27:56,982
want an anti derivative of X
squared and if you don't know

383
00:27:56,982 --> 00:27:59,655
what one is, you refer back to
your table.

384
00:28:00,940 --> 00:28:04,691
An anti derivative of X
squared is X cubed over 3

385
00:28:04,691 --> 00:28:05,714
plus a constant.

386
00:28:14,250 --> 00:28:18,628
So in order to calculate this
definite integral, what we need

387
00:28:18,628 --> 00:28:20,220
to do is evaluate.

388
00:28:21,220 --> 00:28:25,268
The Anti Derivative Capital F.
At B. That's the upper limit,

389
00:28:25,268 --> 00:28:27,476
which in this case is one.

390
00:28:28,720 --> 00:28:32,490
And then at the lower limit,
which in the in our case is

391
00:28:32,490 --> 00:28:34,230
zero. Let's workout F of one.

392
00:28:35,920 --> 00:28:37,864
Well, F of one is going to be 1

393
00:28:37,864 --> 00:28:40,646
cubed. Over 3, which is
just a third.

394
00:28:42,110 --> 00:28:42,790
Plus C.

395
00:28:45,170 --> 00:28:46,928
Let's workout big F of 0.

396
00:28:48,340 --> 00:28:49,560
Big F of 0.

397
00:28:50,620 --> 00:28:54,316
Is going to be 0 cubed over
three, which is 0 plus. See, so

398
00:28:54,316 --> 00:28:57,508
it's just see. And then we
want to find the difference.

399
00:28:58,770 --> 00:29:01,276
Half of 1 minus F of note.

400
00:29:02,620 --> 00:29:06,148
Will be 1/3 plus C minus C, so
the Seas will cancel and would

401
00:29:06,148 --> 00:29:07,660
be just left with the third.

402
00:29:08,890 --> 00:29:11,930
In other words, to calculate
this integral here.

403
00:29:13,280 --> 00:29:16,376
All we have to do is find the
Anti Derivative Capital F.

404
00:29:17,210 --> 00:29:21,236
Evaluate it at the upper limit
evaluated at the lower limit.

405
00:29:21,236 --> 00:29:25,262
Find the difference and the
result is the third. That's the

406
00:29:25,262 --> 00:29:28,190
area under this graph between
North and one.

407
00:29:29,660 --> 00:29:32,169
Now let me just show you how we
would normally set this out.

408
00:29:33,200 --> 00:29:34,904
We would normally set this
out like this.

409
00:29:38,680 --> 00:29:43,918
We find. An anti derivative of X
squared, which we've seen is X

410
00:29:43,918 --> 00:29:45,558
cubed over 3 plus C.

411
00:29:49,300 --> 00:29:52,667
We don't normally write the Plus
C down, and the reason for that

412
00:29:52,667 --> 00:29:55,257
is when we're finding definite
integrals, the sea will always

413
00:29:55,257 --> 00:29:58,624
cancel out as we saw here, the
Seas cancelled out in here and

414
00:29:58,624 --> 00:30:00,178
that will always be the case.

415
00:30:00,380 --> 00:30:03,500
So we don't actually need to
write a plus C down when we

416
00:30:03,500 --> 00:30:05,420
write down an anti
derivative of X squared.

417
00:30:06,590 --> 00:30:10,370
It's conventional to write down
the anti derivative in square

418
00:30:10,370 --> 00:30:14,600
brackets. And to transfer the
limits on the original integral,

419
00:30:14,600 --> 00:30:18,440
the Norton one to the right hand
side over there like so.

420
00:30:20,480 --> 00:30:23,912
What we then want to do is
evaluate the anti derivative at

421
00:30:23,912 --> 00:30:28,202
the top limit that corresponded
to the FB or the F of one. So we

422
00:30:28,202 --> 00:30:31,920
work this out at the top limit
which is 1 cubed over 3.

423
00:30:35,410 --> 00:30:39,232
We work it out at the bottom
limit. That's the F of a or

424
00:30:39,232 --> 00:30:40,324
the F of Norte.

425
00:30:41,380 --> 00:30:44,838
We work this out of the lower
limits will just get zero and

426
00:30:44,838 --> 00:30:46,966
then we want the difference
between the two.

427
00:30:48,430 --> 00:30:51,670
Which is just going to give
us a third. So that's the

428
00:30:51,670 --> 00:30:54,100
normal way we would set out a
definite integral.

429
00:31:00,190 --> 00:31:01,450
So as we've seen.

430
00:31:02,760 --> 00:31:07,208
Definite integration is very
closely associated with Anti

431
00:31:07,208 --> 00:31:07,764
Differentiation.

432
00:31:10,190 --> 00:31:14,326
Because of this, in general it's
useful to think of Anti

433
00:31:14,326 --> 00:31:18,086
Differentiation as integration
and then we would often refer to

434
00:31:18,086 --> 00:31:21,846
a table like this one of
antiderivatives as simply a

435
00:31:21,846 --> 00:31:22,974
table of integrals.

436
00:31:24,020 --> 00:31:27,153
And there will be a table of
integrals in the notes and in

437
00:31:27,153 --> 00:31:30,045
later videos you'll be looking
in much more detail at how to

438
00:31:30,045 --> 00:31:32,937
use a table of integrals. So
you think of the table of

439
00:31:32,937 --> 00:31:35,106
integrals as your table of
antiderivatives and vice versa.

440
00:31:36,730 --> 00:31:40,258
Very early on in the video, we
looked at this problem. We

441
00:31:40,258 --> 00:31:44,080
started off with little F of X
being equal to four X cubed.

442
00:31:45,350 --> 00:31:47,198
Minus Seven X squared.

443
00:31:48,280 --> 00:31:51,310
Plus 12X minus 4.

444
00:31:52,390 --> 00:31:53,740
We differentiate it.

445
00:31:57,250 --> 00:32:02,092
To give 12 X squared
minus 14X plus 12.

446
00:32:03,140 --> 00:32:08,249
And then we said that an anti
derivative of 12 X squared minus

447
00:32:08,249 --> 00:32:09,428
14X plus 12.

448
00:32:10,040 --> 00:32:12,644
Was this function
capital F plus C?

449
00:32:13,840 --> 00:32:17,400
We've got a notation we
can use now because we've

450
00:32:17,400 --> 00:32:19,892
linked antiderivatives to
integrals and the notation

451
00:32:19,892 --> 00:32:23,808
we would use is that the
integral of 12 X squared.

452
00:32:25,260 --> 00:32:31,188
Minus 14X plus 12 DX
is equal to.

453
00:32:31,890 --> 00:32:33,250
4X cubed

454
00:32:34,410 --> 00:32:40,180
minus Seven X squared plus 12X
plus any arbitrary constant.

455
00:32:41,050 --> 00:32:43,675
And we call this an
indefinite integral.

456
00:32:50,120 --> 00:32:54,680
And we say that the indefinite
integral of 12 X squared minus

457
00:32:54,680 --> 00:33:00,000
14X plus 12 with respect to X is
4X cubed minus Seven X squared

458
00:33:00,000 --> 00:33:02,740
plus 12X. Plus a constant
of integration.

459
00:33:06,520 --> 00:33:10,570
OK, In summary, what have
we found? What we found

460
00:33:10,570 --> 00:33:12,190
that a definite integral?

461
00:33:14,090 --> 00:33:20,075
The integral from A to B of
little F of X DX. We found that

462
00:33:20,075 --> 00:33:21,671
this is a number.

463
00:33:22,530 --> 00:33:28,052
And it's obtained from the
formula F of B minus FA,

464
00:33:28,052 --> 00:33:33,072
where big F is any anti
derivative of Little F.

465
00:33:34,500 --> 00:33:38,451
And we've also seen the
indefinite integral of F

466
00:33:38,451 --> 00:33:39,768
of X DX.

467
00:33:40,880 --> 00:33:46,496
Is a function big F of X plus an
arbitrary constant? Again, where

468
00:33:46,496 --> 00:33:51,248
F is any anti derivative of
Little F&C is an arbitrary

469
00:33:51,248 --> 00:33:56,059
constant. So in this video we've
learned how to do

470
00:33:56,059 --> 00:33:58,585
differentiation in reverse.
We've learned about

471
00:33:58,585 --> 00:34:00,480
antiderivatives. Definite

472
00:34:00,480 --> 00:34:04,134
integrals. And indefinite
integrals in subsequent videos.

473
00:34:04,134 --> 00:34:07,770
You learn a lot more teak
techniques of integration.